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Question Number 6482 by prakash jain last updated on 28/Jun/16

Is there f:R→R such that f(x)≠0 f(x^3 )=3f(x^2 )

$$\mathrm{Is}\:\mathrm{there}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:{f}\left({x}\right)\neq\mathrm{0} \\ $$$${f}\left({x}^{\mathrm{3}} \right)=\mathrm{3}{f}\left({x}^{\mathrm{2}} \right) \\ $$$$ \\ $$

Commented by Yozzii last updated on 29/Jun/16

f(x^3 )=3f(x^2 ) x=0⇒f(0)=3f(0)⇒f(0)=0. x=1⇒f(1)=3f(1)⇒f(1)=0 x=−1⇒f(−1)=3f(1)=0 x=0,±1 are some roots of f(x)=0, if ever f(x)=0 as f(x) is not everywhere zero. f(x)=Σ_(i=0) ^∞ a(i)x^i =a(0)+a(1)x+a(2)x^2 +a(3)x^3 +... {a(i)}_(i=0) ^∞ is a real sequence 3f(x^2 )=3a(0)+3a(1)x^2 +3a(2)x^4 +3a(3)x^6 +3a(4)x^8 +3a(5)x^(10) +3a(6)x^(12) ... f(x^3 )=a(0)+a(1)x^3 +a(2)x^6 +a(3)x^9 +a(4)x^(12) +... f(x^3 )=3f(x^2 ) ⇒a(i)=0 for all i∈Z^≥ ⇒f(x)=0. ⇒f(x)≠Σ_(i=0) ^∞ a(i)x^i

$${f}\left({x}^{\mathrm{3}} \right)=\mathrm{3}{f}\left({x}^{\mathrm{2}} \right) \\ $$$${x}=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{3}{f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}. \\ $$$${x}=\mathrm{1}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{3}{f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1}\Rightarrow{f}\left(−\mathrm{1}\right)=\mathrm{3}{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0},\pm\mathrm{1}\:{are}\:{some}\:{roots}\:{of}\:{f}\left({x}\right)=\mathrm{0},\:{if}\:{ever} \\ $$$${f}\left({x}\right)=\mathrm{0}\:{as}\:{f}\left({x}\right)\:{is}\:{not}\:{everywhere}\:{zero}. \\ $$$${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}\left({i}\right){x}^{{i}} ={a}\left(\mathrm{0}\right)+{a}\left(\mathrm{1}\right){x}+{a}\left(\mathrm{2}\right){x}^{\mathrm{2}} +{a}\left(\mathrm{3}\right){x}^{\mathrm{3}} +... \\ $$$$\left\{{a}\left({i}\right)\right\}_{{i}=\mathrm{0}} ^{\infty} \:{is}\:{a}\:{real}\:{sequence} \\ $$$$\mathrm{3}{f}\left({x}^{\mathrm{2}} \right)=\mathrm{3}{a}\left(\mathrm{0}\right)+\mathrm{3}{a}\left(\mathrm{1}\right){x}^{\mathrm{2}} +\mathrm{3}{a}\left(\mathrm{2}\right){x}^{\mathrm{4}} +\mathrm{3}{a}\left(\mathrm{3}\right){x}^{\mathrm{6}} +\mathrm{3}{a}\left(\mathrm{4}\right){x}^{\mathrm{8}} +\mathrm{3}{a}\left(\mathrm{5}\right){x}^{\mathrm{10}} +\mathrm{3}{a}\left(\mathrm{6}\right){x}^{\mathrm{12}} ... \\ $$$${f}\left({x}^{\mathrm{3}} \right)={a}\left(\mathrm{0}\right)+{a}\left(\mathrm{1}\right){x}^{\mathrm{3}} +{a}\left(\mathrm{2}\right){x}^{\mathrm{6}} +{a}\left(\mathrm{3}\right){x}^{\mathrm{9}} +{a}\left(\mathrm{4}\right){x}^{\mathrm{12}} +... \\ $$$${f}\left({x}^{\mathrm{3}} \right)=\mathrm{3}{f}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{a}\left({i}\right)=\mathrm{0}\:{for}\:{all}\:{i}\in\mathbb{Z}^{\geqslant} \Rightarrow{f}\left({x}\right)=\mathrm{0}. \\ $$$$\Rightarrow{f}\left({x}\right)\neq\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}\left({i}\right){x}^{{i}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by nburiburu last updated on 29/Jun/16

f(x^k )= ((∂^k (x^k ))/((∂x)^k )) . f(x^3 )=(x^3 )′′′=(3x^2 )′′=(6x)′=6 f(x^2 )=(x^2 )′′=(2x)′=2 f(x^3 )=3f(x^2 )

$${f}\left({x}^{{k}} \right)=\:\frac{\partial^{{k}} \left({x}^{{k}} \right)}{\left(\partial{x}\right)^{{k}} }\:. \\ $$$${f}\left({x}^{\mathrm{3}} \right)=\left({x}^{\mathrm{3}} \right)'''=\left(\mathrm{3}{x}^{\mathrm{2}} \right)''=\left(\mathrm{6}{x}\right)'=\mathrm{6} \\ $$$${f}\left({x}^{\mathrm{2}} \right)=\left({x}^{\mathrm{2}} \right)''=\left(\mathrm{2}{x}\right)'=\mathrm{2} \\ $$$${f}\left({x}^{\mathrm{3}} \right)=\mathrm{3}{f}\left({x}^{\mathrm{2}} \right) \\ $$$$ \\ $$

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